Limit of a function | IIT JAM 2017 | Problem 8

Given that, $ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

therefore,

$ f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

$ f(x)=\bigg\{\begin{array}{cc}
(2+x) \sin \left(\frac{1}{x}\right), & , x>0 \\
-x \sin \left(\frac{1}{x}\right), & x<0 \\
\end{array} $

Let, $L=\displaystyle\lim_{x \to 0^{-}} f(x)$

and , $R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .$

Now,

$L= \displaystyle\lim_{x \to 0^{-}} f(x) $

$\quad = \displaystyle\lim_{x \to 0^{-}} -x \sin \left(\frac{1}{x}\right) $

$ \quad = -\displaystyle\lim_{x \to 0^{-}} x \sin \left(\frac{1}{x}\right) $

Theorem : If $D \subset \mathbb R$ and $f,g : D \to \mathbb R$ . Let $c \in D$. If f is bounded on $N'(c)\cup D$ and $\displaystyle\lim_{x \to c} g(x)=0$, then $\displaystyle\lim_{x \to c}(f.g)(x)=0$.

Now , $ \sin \left(\frac{1}{x}\right) $ is bounded in $\mathbb R - \{0\}$ and $ \displaystyle\lim_{x \to 0^{-}} x=0$ , then $\displaystyle\lim_{x \to 0^{-}} f(x)$ exists and equal to $0$.

But,

$R=\displaystyle\lim_{x\to 0^{+}}f(x)$

$\quad = \displaystyle\lim_{x\to 0^{+}} (2+x) \sin \left(\frac{1}{x}\right),$

$\quad= \displaystyle\lim_{x\to 0^{+}} 2\sin \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right) $

$\lim_{x \to 0^{+}} \sin \left(\frac{1}{x}\right) $ does not exists [Why?]

Then $L$ exists but $R$ does not.

Given that, $ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

therefore,

$ f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

$ f(x)=\bigg\{\begin{array}{cc}
(2+x) \sin \left(\frac{1}{x}\right), & , x>0 \\
-x \sin \left(\frac{1}{x}\right), & x<0 \\
\end{array} $

Let, $L=\displaystyle\lim_{x \to 0^{-}} f(x)$

and , $R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .$

Now,

$L= \displaystyle\lim_{x \to 0^{-}} f(x) $

$\quad = \displaystyle\lim_{x \to 0^{-}} -x \sin \left(\frac{1}{x}\right) $

$ \quad = -\displaystyle\lim_{x \to 0^{-}} x \sin \left(\frac{1}{x}\right) $

Theorem : If $D \subset \mathbb R$ and $f,g : D \to \mathbb R$ . Let $c \in D$. If f is bounded on $N'(c)\cup D$ and $\displaystyle\lim_{x \to c} g(x)=0$, then $\displaystyle\lim_{x \to c}(f.g)(x)=0$.

Now , $ \sin \left(\frac{1}{x}\right) $ is bounded in $\mathbb R - \{0\}$ and $ \displaystyle\lim_{x \to 0^{-}} x=0$ , then $\displaystyle\lim_{x \to 0^{-}} f(x)$ exists and equal to $0$.

But,

$R=\displaystyle\lim_{x\to 0^{+}}f(x)$

$\quad = \displaystyle\lim_{x\to 0^{+}} (2+x) \sin \left(\frac{1}{x}\right),$

$\quad= \displaystyle\lim_{x\to 0^{+}} 2\sin \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right) $

$\lim_{x \to 0^{+}} \sin \left(\frac{1}{x}\right) $ does not exists [Why?]

Then $L$ exists but $R$ does not.